Problem: Plot $(-6, 10)$ and select the quadrant in which the point lies. $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $y$ $x$ $ \text{ QI }$ $ \text{ QII }$ $ \text{ QIII }$ $ \text{ QIV }$
Coordinates are listed as $({x},{y})$ So, for $( {-6}, {10} )$ our $x$ -coordinate is ${-6}$ and our $y$ -coordinate is ${10}$ The $x$ -coordinate tells how far we move to the right from the origin and the $y$ -coordinate tells us how far we move up from the origin. Since our $x$ -coordinate is negative, we move ${6}$ to the left. Since our $y$ -coordinate is positive, we move ${10}$ up. Move the point to $( {-6}, {10} )$ at the marked point above. Now that we have our point plotted, we can figure out the quadrant. By convention, quadrants are named with a capital $\text{Q}$ and a roman numeral, starting in the upper right quadrant as $\text{QI}$ and rotating counter-clockwise. Since our point is in the upper left portion of the graph, the quadrant is ${\text{QII}}$.